Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#define _USE_MATH_DEFINES
 
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
 
using namespace std;
 
typedef pair<long long int, long long int> P;
long long int INF = 1e18;

#define MAX_V 110000

int N;

struct edge{
	int to;
	long long int cost;
};

vector<edge> G[MAX_V];
long long int d[MAX_V];

// 速さはO(|E|log|V|)

void shortest_path(int s, int V){ // Vは頂点数
	
	priority_queue<P, vector<P>, greater<P> > que;
	
	for(int i = 0; i < V; i++){
		d[i] = INF;
	}
	d[s] = 0;
	
	que.push(P(0, s));
	
	while(!que.empty()){
		
		P p = que.top();
		que.pop();
		
		int v = p.second;
		
		if(d[v] < p.first){
			continue;
		}
		
		for(int i = 0; i < G[v].size(); i++){
			edge e = G[v][i];
			if(d[e.to] > e.cost + d[v]){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		} // 頂点への最小コストを更新
	}
}

long long int cnt1[310000] = {}, cnt2[310000] = {};

int main(){
	
	long long int N, M, s, t;
	cin >> N >> M >> s >> t;
	s--;
	t--;
	
	for(int i = 0; i < M; i++){
		edge e;
		e.cost = 1;
		int u, v;
		cin >> u >> v;
		u--;
		v--;
		e.to = v;
		G[u].push_back(e);
		e.to = u;
		G[v].push_back(e);
	}
	
	shortest_path(s, N);
	
	for(int i = 0; i < N; i++){
		if(d[i] <= M){
			cnt1[d[i]]++;
		}
	}
	
	int dist = d[t];
	
	shortest_path(t, N);
	
	for(int i = 0; i < N; i++){
		if(d[i] <= M){
			cnt2[d[i]]++;
		}
	}
	
	long long int ans = 0;
	
	for(int i = 0; i <= dist - 2; i++){
		ans += cnt1[i] * cnt2[dist - i - 2];
	}
	
	cout << ans << endl;
	
	return 0;
}